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I cant follow the logic of this proof

  1. Mar 6, 2009 #1
    W_1 and W_ 2 are subspaces of V of inner product V.
    prove that if
    [tex]
    W_1\subseteq W_2
    [/tex]
    then
    [tex]
    W_1^\perp \supseteq W_2^\perp
    [/tex]

    the proof is:
    we take [tex]v\epsilon W_1[/tex]
    so <v,w>=0 for every [tex]w\epsilon W_1^\perp [/tex]

    and because
    [tex]
    W_2^\perp \subseteq W_1^\perp
    [/tex]
    (i cant see why the "viven expression is "true" why "because")
    we get that
    <v,w>=0
    for every [tex]w\epsilon W_2^\perp [/tex]

    so [tex]v\epsilon W_2[/tex]
    so
    [tex]
    W_1^\perp \supseteq W_2^\perp
    [/tex]

    ??
     
    Last edited: Mar 6, 2009
  2. jcsd
  3. Mar 6, 2009 #2

    Office_Shredder

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    Gold Member

    Your latex is messed up a bit, but it looks like what you typed after "because" is intended to be the final conclusion, which doesn't make a whole lot of sense. What you want is

    <v,w>=0 for all w in W2 then <v,w> = 0 for all w in W1 (because W1 is a subset) Hence v in W2perp implies v is in W1perp by the definition of the perpendicular subspace
     
  4. Mar 6, 2009 #3
    sorry i ment
    [tex]

    W_2^\perp \subseteq W_1^\perp

    [/tex]


    i cant see how they came to this conclusion after they defined v
    ??
     
  5. Mar 6, 2009 #4
    we got two subspaces W1 and W2
    W2 includes W1
    they take "v" which is orthogonal to every vector which is orthogonal to W1

    so v is parallel to W1 ,v is a part of W1 .

    you say
    "<v,w>=0 for all w in W2"
    so all the vectors in w (W2) are perpendicular to v (which is W1)

    no one told us that W2 is perpendicular to W1 ??

    further more i was told that W2 include W1

    W2 W1 cannot be orthogonal
    ??
     
  6. Mar 6, 2009 #5

    HallsofIvy

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    Staff Emeritus
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    The way to prove A is a subset of B is to start "if x is a member of A" and conclude "x is a member of B".
    If x is a member of [tex]W_2^\perp[/tex] then <x, v>= 0 for every member, v, of [tex]W_2[/tex]. Since [tex]W_1[/tex] is a subset of [tex]W_2[/tex], every v in [tex]W_1[/tex] is a member of [tex]W_2[/tex] so <x, v>= 0 for every member of [tex]W_1[/tex].
     
  7. Mar 6, 2009 #6
    ok i agree <v,x>=0 because W1 is a part of W2

    what is the next logical step
    its still doesnt give me the step i didnt understang

    [tex]
    W_2^\perp \subseteq W_1^\perp
    [/tex]
    how did they get it?
     
  8. Mar 6, 2009 #7
    v is perpendicular to [tex] W_1^\perp [/tex]
    v is perpendicular to [tex] W_2^\perp [/tex] because W1 is a part of W2
    so
    one perpendicular member is a part of the other perpendicular member
    but how to decide W2 perpendicular is a part of W1 perpendicular

    [tex]
    W_2^\perp \subseteq W_1^\perp
    [/tex]
    why its not the other way around
    ??
     
  9. Mar 7, 2009 #8
    how to deside?
     
  10. Mar 7, 2009 #9

    Mark44

    Staff: Mentor

    Maybe a diagram would help. Here's a representation of the vector space V and the two subspaces, [itex]W_1 and W_2[/itex], with [itex]W_1 \subseteq W_2[/itex].



    [itex]W_2^\perp[/itex] is everything outside of [itex]W_2[/itex], but including the 0 vector.
    [itex]W_1^\perp[/itex] is everything outside of [itex]W_1[/itex], which means it includes everything in [itex]W_2[/itex] that's not also in [itex]W_1[/itex], plus everything that's outside of [itex]W_2[/itex], plus the 0 vector.

    Now, any vector [itex]v \in W_1^\perp[/itex] must be in
    [itex](W_2 - W_1) \cup \{0\} \cup W_2^\perp[/itex]

    [itex]\forall v \in W_2^\perp[/itex],
    [itex]v \in W_2^\perp \cup (W_2 - W_1) \cup \{0\} = W_1^\perp[/itex].
    Therefore, [itex] W_2^\perp \subseteq W_1^\perp [/itex]
     

    Attached Files:

  11. Mar 7, 2009 #10
  12. Mar 7, 2009 #11

    arildno

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    We know that every perpendicular vector of the larger subspace W_2 must be perpendicular to all vectors of any subspace within W_2; including the vectors making up W_1.

    Thus, ALL vectors in the perpendicularity space of W_2, are elements in the perpendicularity space of W_1.
    This means that the perp space of W_2 is contained in that of W_1

    But, does it thereby follow that the perpendicularity space of W_1 is exhausted?
    NO!

    It might well include vectors of W_2 that happens to be perpendicular to every vector in W_1 as well!

    Such vectors would NOT be part of the perpindicularity space of W_2, and therefore, one cannot say that the perp space of W_1 is contained within that of W_2
     
  13. Mar 7, 2009 #12
    i cant see the attached diagram
     
  14. Mar 7, 2009 #13
    "Thus, ALL vectors in the perpendicularity space of W_2, are elements in the perpendicularity space of W_1."
    why its not the other way around
     
  15. Mar 7, 2009 #14

    arildno

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    Because a non-zero vector in W_2 that is not in W_1 might be perpendicular to all vectors in W_1.

    Since it by definition would be IN W_2, it cannot be perpendicular to every vector in W_2 (it is, for example, NOT perpendicular to itself!).

    Thus, such a vector would be in the perpendicularity space of W_1, but not in the perpendicularity space of W_2
     
  16. Mar 7, 2009 #15

    Mark44

    Staff: Mentor

    You should be able to see it now.
     
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